Rational Sequences and Dominant Powers

Why Dominant Powers Determine the Limit

Polynomial expressions can contain several powers of \(n\), but as \(n\) grows, the highest power eventually matters most. For a quotient of two polynomials, the comparison between their highest powers determines whether the quotient approaches zero, approaches a nonzero constant, or grows without bound.

This makes rational sequences a natural next step after polynomial sequence limits. Instead of separately estimating every term, we identify the dominant power in the numerator and denominator and compare them carefully.

Central idea: For a rational sequence, compare the numerator degree with the denominator degree. If the degrees are equal, the limit is the ratio of the leading coefficients.

Degrees and Leading Coefficients

A nonzero polynomial \(p(n)\) of degree \(m\) can be written as

p(n)=a_m n^m+a_{m-1}n^{m-1}+\cdots+a_1n+a_0, \qquad a_m\ne 0.

The degree of \(p\) is \(m\), and \(a_m\) is its leading coefficient. The term \(a_m n^m\) is the polynomial's dominant term as \(n\) tends to infinity. For example, \(4n^3-2n+7\) has degree \(3\) and leading coefficient \(4\).

Consider a rational sequence whose terms are given by \(p(n)/q(n)\), where \(p\) and \(q\) are nonzero polynomials. We assume the quotient is defined for all sufficiently large positive integers \(n\). This condition is enough for a limit: changing or omitting finitely many initial terms does not change the limit of a sequence.

Definition: If \(p\) has degree \(m\) and leading coefficient \(a_m\), and \(q\) has degree \(k\) and leading coefficient \(b_k\), then \(m\) and \(k\) are the numerator and denominator degrees, while \(a_m\) and \(b_k\) are their leading coefficients.

The Degree-Comparison Rule

There are three possibilities. A smaller numerator degree produces a zero limit. Equal degrees produce a finite limit given by the ratio of leading coefficients. A larger numerator degree produces terms whose magnitudes grow without bound.

Theorem: Let \(p\) and \(q\) be nonzero polynomials of degrees \(m\) and \(k\), with leading coefficients \(a_m\) and \(b_k\), respectively, where \(b_k\ne 0\). Then, as \(n\to\infty\) through positive integers:

If \(m<k\), then \(\displaystyle \frac{p(n)}{q(n)}\to 0\).
If \(m=k\), then \(\displaystyle \frac{p(n)}{q(n)}\to \frac{a_m}{b_k}\).
If \(m>k\), the quotient has no finite real limit. Its magnitude tends to infinity; its sign is eventually the sign of \(a_m/b_k\).

Why the Rule Works

Divide the numerator by \(n^m\) and the denominator by \(n^k\). The terms below the highest power become powers of \(1/n\), which tend to zero. This is the key effect of dominant powers.

In detail, if \(p(n)=a_m n^m+\cdots+a_0\), then

\frac{p(n)}{n^m} = a_m+\frac{a_{m-1}}{n}+\frac{a_{m-2}}{n^2} +\cdots+\frac{a_0}{n^m} \longrightarrow a_m.

Likewise, \(q(n)/n^k\to b_k\). The sequence limit laws established in the earlier material on sum, product, and quotient limits allow us to take quotients when the denominator limit is nonzero. The exact identity connecting these normalized polynomials to the original quotient is

\frac{p(n)}{q(n)} = n^{m-k} \frac{p(n)/n^m}{q(n)/n^k}.

The fraction on the right tends to \(a_m/b_k\), a nonzero number. The remaining factor \(n^{m-k}\) determines the three cases: it tends to zero if \(m<k\), equals \(1\) if \(m=k\), and grows without bound if \(m>k\). When \(m>k\), the factor is positive for positive \(n\), so the eventual sign is the sign of \(a_m/b_k\).

Find the degrees.
Identify the highest power of \(n\) in each polynomial.

Compare the degrees.
Determine whether the numerator degree is lower, equal, or higher.

Apply the matching case.
For equal degrees, divide the leading coefficients; for a higher numerator degree, also determine the eventual sign.

Worked Examples

Worked Example: A Lower-Degree Numerator

Find the limit of

a_n=\frac{5n+2}{2n^3-3n+1}.

The numerator has degree \(1\) and leading coefficient \(5\). The denominator has degree \(3\) and leading coefficient \(2\). Since \(1<3\), the degree-comparison rule predicts a limit of zero.

To verify it directly, divide the numerator and denominator by \(n^3\):

a_n = \frac{\frac{5}{n^2}+\frac{2}{n^3}} {2-\frac{3}{n^2}+\frac{1}{n^3}}.

As \(n\to\infty\), the numerator tends to \(0+0=0\), and the denominator tends to \(2-0+0=2\). Since \(2\ne0\), the quotient limit law applies, giving

\lim_{n\to\infty}a_n=\frac{0}{2}=0.

Check: The denominator's degree exceeds the numerator's by \(2\), and the normalized expression contains only terms with negative powers in its numerator. The computed limit \(0\) agrees with that degree comparison.

Worked Example: Equal Degrees

Find the limit of

b_n=\frac{3n^2-2n+5}{6n^2+4n-1}.

Both polynomials have degree \(2\). Their leading coefficients are \(3\) and \(6\), so the predicted limit is \(3/6=1/2\). Divide the numerator and denominator by \(n^2\) to check:

b_n = \frac{3-\frac{2}{n}+\frac{5}{n^2}} {6+\frac{4}{n}-\frac{1}{n^2}}.

The numerator tends to \(3-0+0=3\), and the denominator tends to \(6+0-0=6\). Its limit is nonzero, so

\lim_{n\to\infty}b_n=\frac{3}{6}=\frac12.

Check: Dividing the original numerator and denominator by \(n^2\) leaves leading terms \(3\) and \(6\); all other terms tend to zero. Thus the result is exactly the ratio of the leading coefficients.

Worked Example: A Higher-Degree Numerator

Determine the behavior as \(n\to\infty\) of

c_n=\frac{-2n^3+5n-1}{n^2+4}.

The numerator has degree \(3\) with leading coefficient \(-2\). The denominator has degree \(2\) with leading coefficient \(1\). The numerator degree is larger, and the ratio of leading coefficients is \(-2/1=-2\), which is negative. The rule predicts that the sequence tends to \(-\infty\).

For verification, factor out the dominant powers by dividing the numerator by \(n^3\) and the denominator by \(n^2\):

c_n = n\, \frac{-2+\frac{5}{n^2}-\frac{1}{n^3}} {1+\frac{4}{n^2}}.

The fraction tends to \(-2/1=-2\). In particular, for all sufficiently large \(n\), it is negative and bounded away from zero. Multiplication by \(n\), which grows without bound and remains positive, therefore gives

c_n\longrightarrow-\infty.

Check: The leading powers give \((-2n^3)/(n^2)=-2n\), which is negative and unbounded in magnitude. This confirms both the sign and the unbounded behavior.

A Common Mistake: Keeping Only the Leading Coefficients

Leading coefficients determine the limit only when the degrees are equal. For example, it would be incorrect to claim that \((5n+2)/(2n^3-3n+1)\) tends to \(5/2\) just because \(5\) and \(2\) are the leading coefficients. The powers attached to those coefficients differ, so the coefficient ratio alone is not the answer.

The incorrect calculation effectively treats \(5n\) and \(2n^3\) as if they had the same power. The correction is to compare degrees first. Here the numerator degree is \(1\) and the denominator degree is \(3\), so the limit is \(0\), as the first worked example showed. A reliable order is: compare degrees; only if they match, take the ratio of leading coefficients.

What to Conclude in Each Case

Degree comparison	Limit behavior	What determines the result?
Numerator degree is smaller	Limit is \(0\)	The denominator grows by a higher power.
Degrees are equal	Finite nonzero limit	Ratio of leading coefficients.
Numerator degree is larger	No finite real limit; magnitude grows without bound	Degree difference gives the growing power; leading-coefficient ratio gives the eventual sign.

The equal-degree case is the one in which the leading-coefficient ratio is itself the limit. In the higher-degree case, it helps determine the sign, but the limit is not that ratio: the remaining positive power of \(n\) makes the magnitude unbounded. In the lower-degree case, the limit is zero regardless of the particular nonzero leading coefficients.

Check Your Understanding

For each sequence, identify the numerator and denominator degrees before deciding its limit or behavior.

What is the limit of \(\displaystyle \frac{7n^2+n}{3n^4+2}\)?
What is the limit of \(\displaystyle \frac{8n^3-1}{4n^3+5n}\)?
Does \(\displaystyle \frac{n^4+2}{-3n^2+1}\) have a finite real limit? If not, what is its eventual sign?
In which degree case is the limit the ratio of the leading coefficients?
Why is it not enough to compare leading coefficients before comparing degrees?